Question 1031682
What you do with
{{{1/(x+1) + 1/(x+2) = 1 }}}
is multiply everything by the lowest common denominator, (x+1)(x+2)...we get
{{{(x+2) + (x+1) = (x+1)(x+2)}}}
Now solve for x...we get
{{{2x + 3 = x^2 + 3x + 2}}}
{{{x^2 + 2x - 1 = 0}}}
which cannot be factored...you'll have to apply the quadratic formula from here...